Recently, energy-saving has been actively promoted in terms of environmental measure. For portable equipments using battery such as a mobile phone, a digital camera and so on, it is especially important to have a longer battery life. Such portable equipment commonly uses a constant voltage circuit which supplies a low voltage power. The constant voltage circuit generally includes a boost circuit to obtain the low voltage power with a full voltage range.
FIG. 1 illustrates a conventional boost circuit. The conventional boost circuit 100 includes an inductor L101, a switching device M101 and a diode D101. The switching device M101 is formed of a NMOS (n-type metal oxide semiconductor) transistor and is connected between an input terminal IN and a ground through the inductor L101. The diode D101 works as a rectifying device and is connected between a node at which the inductor L101 is wired with the switching device M101 and an output terminal OUT.
In this conventional boost circuit, electric energy is stored in the inductor L101 while the switching device M101 is on. When the switching device M101 is off, the electric energy stored in the inductor L101 is output by being added with an input voltage Vin. However, a power conversion efficiency of this conventional boost circuit is not high because there is a relatively large voltage drop at the diode D101.
FIG. 2 illustrates another conventional boost circuit 200. The boost circuit 200 employs a PMOS (p-type metal oxide semiconductor) transistor M102 as a rectifying device instead of the diode D101. The switching device M101 and the PMOS transistor M102 are switched to be on and off complementarily. A voltage drop of the PMOS transistor M102 is relatively small when the PMOS transistor M102 is on in comparison to the voltage drop of the diode D101 in FIG. 1. As a result, the power conversion efficiency is much improved using the PMOS transistor M102 as the rectifying device.
A substrate gate of the PMOS transistor M102 is wired to a node of the output terminal side of the PMOS transistor M102. The parasitic diode D102 is formed between a node of the inductor L101 side of the PMOS transistor M102 and the substrate gate of the PMOS transistor M102. An anode of the parasitic diode D102 is wired to the node of the inductor L101 side of the PMOS transistor M102, and a cathode of the parasitic diode D102 is wired to the node of the output terminal OUT side of the PMOS transistor M102.
In a suspend mode of the boost circuits, the input voltage Vin is output through the diode D101 in the boost circuit 100 of FIG. 1. And the input voltage Vin is output to the output terminal OUT through the parasitic diode D102 in the boost circuit 200 of FIG. 2. Thus, The power is continuously being supplied and cannot be stopped even if the boost circuit is suspended.
FIG. 3 illustrates another conventional boost circuit 300. The boost circuit 300 includes two rectifying devices M112 and M113. Each rectifying device M112 and M113 includes parasitic diode D112 and D113. A pair of the rectifying device and parasitic diode are wired in series so that polarities of the pair of the rectifying device and parasitic diode are arranged in reverse direction from each other as shown in FIG. 3. The switching device M111, the rectifying devices M112 and M113 are set to be off by a control circuit 113 in accordance with a feedback signal from an overload detector which detects an overload condition and a protection circuit. As a result, an input voltage Vin is prevented from being output to the output terminal OUT.
However, the boost circuit 300 needs two larger rectifying devices because a large overload current flows through the rectifying devices. As a result, there is a cost penalty due to a necessity of large devices in size when the circuit is fabricated onto an IC (integrated circuit). Further, a power conversion efficiency is decreased in the boost circuit 300 because of a voltage drop at the rectifying devices M112 and M113 which is twice larger than the other boost circuits such as the boost circuit 200.